giải phương trình sau : (149-x)/25+(170-x)/23+(187-x)/21+(200-x)/19=10
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H
2
20 tháng 4 2018
Ta có : \(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)
\(\Leftrightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)
\(\Leftrightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)
\(\Leftrightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Vì \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\)
Nên : 124 - x = 0
<=> x = 124
Vậy x = 124
VQ
3
NH
29 tháng 3 2023
\(\dfrac{x-187}{13}+\dfrac{x-170}{15}+\dfrac{x-149}{17}+\dfrac{x-124}{19}=10\)
`<=>(x-187)/13+(x-170)/15+(x-149)/17+(x-124)/19-10=0`
`<=>(x-187)/13-1+(x-170)/15-2+(x-149)/17-3+(x-124)/19-4=0`
`<=>(x-200)/13+(x-200)/15+(x-200)/17+(x-200)/19=0`
`<=>(x-200)(1/13+1/15+1/17+1/19)=0`
`<=>x-200=0(1/13+1/15+1/17+1/19>0)`
`<=>x=200`
29 tháng 3 2023
\(=>\left(\dfrac{x-187}{13}-1\right)+\left(\dfrac{x-170}{15}-2\right)+\left(\dfrac{x-149}{17}-3\right)+\left(\dfrac{x-124}{19}-4\right)=0\)\(< =>\left(\dfrac{x-187}{13}-\dfrac{13}{13}\right)+\left(\dfrac{x-170}{15}-\dfrac{30}{15}\right)+\left(\dfrac{x-149}{17}-\dfrac{51}{17}\right)+\left(\dfrac{x-124}{19}-\dfrac{76}{19}\right)=0\)
\(< =>\left(\dfrac{x-200}{13}\right)+\left(\dfrac{x-200}{15}\right)+\left(\dfrac{x-200}{17}\right)+\left(\dfrac{x-200}{19}\right)=0\)
\(< =>\left(x-200\right)\left(\dfrac{1}{13}+\dfrac{1}{15}+\dfrac{1}{17}+\dfrac{1}{19}\right)=0\)
\(< =>x-200=0\)
<=>x=200
MD
4
LA
27 tháng 3 2018
\(\frac{148-x}{25}+\frac{179-x}{23}+\frac{206-x}{21}+\frac{229-x}{19}=10\)
\(\Leftrightarrow\frac{148-x}{25}-1+\frac{179-x}{23}-2+\frac{206-x}{21}-3+\frac{229-x}{19}-4=0\)
\(\Leftrightarrow\frac{148-25-x}{25}+\frac{179-46-x}{23}+\frac{206-63-x}{21}+\frac{229-76-x}{19}=0\)
\(\Leftrightarrow\frac{123-x}{25}+\frac{133-x}{23}+\frac{143-x}{21}+\frac{153-x}{19}=0\)
Tới đây bn tự làm tiếp
19 tháng 2 2019
\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+\left(\frac{x}{2003}-1\right)\)
\(\Leftrightarrow\frac{2-x+2001}{2001}=\frac{1-x+2002}{2002}+\frac{x-2003}{2003}\)
\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{x-2003}{2003}\)
\(\Leftrightarrow\left(x-2003\right)\left(\frac{1}{2003}+\frac{1}{2001}-\frac{1}{2002}\right)=0\)
\(\Leftrightarrow x-2003=0\)\(\left(v\text{ì}\frac{1}{2003}+\frac{1}{2001}-\frac{1}{2002}\ne0\right)\)
\(\Leftrightarrow x=2003\)
Vậy \(S=\left\{2003\right\}\)
HN
19 tháng 2 2019
d)Ta có : \(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1-2=\frac{1-x}{2002}+1+1-\frac{x}{2003}-2\)\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\Leftrightarrow\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\)\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow2003-x=0\Leftrightarrow x=2003\)
Vậy phương trình có tập nghiệm S = { 2003 }
LY
26 tháng 3 2017
E = ( x - 29 ) / 1970 + ( x - 27 ) / 1972 + ( x - 25 ) / 1974 + ( x - 23 ) / 1976 + ( x - 21 ) / 1978 + ( x - 19 ) / 1980 = ( x - 1970 ) / 29 + ( x - 1972 ) / 27 + ( x - 1974 ) / 25 + ( x - 1976 ) / 23 + ( x - 1978 ) / 21 + ( x - 1980 ) / 19
( Trừ từng số hạng cho 1 ra như sau )
E = (x - 1999)/ 1970 + ( x - 1999 ) / 1972 + ( x - 1999) / 1974 + ( x - 1999)/ 1976 + ( x -1999) / 1978 + ( x - 1999)/ 1980 = ( x - 1999)/29 + ( x - 1999) / 27 + ( x - 1999 ) / 25 + ( x - 1999) / 23 + ( x - 1999)/21 + ( x - 1999) / 19
< = > ( x - 1999 ) / 1970 + ( x - 1999 ) / 1972 + ( x - 1999 ) / 1974 + ( x - 1999) / 1976 + ( x - 1999) / 1978 + ( x - 1999) / 1980 - ( x - 1999) / 29 - ( x - 1999)/ 27 - ( 1 - 1999) / 25 - ( x-1999) / 23 - ( x - 1999) / 21 - ( x - 1999) / 19 = 0 ( chuyển vế )
< = > ( x - 1999 ) ( 1/1970 + 1/ 1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 - 1/27 - 1/25 - 1/23 - 1/21 - 1/19) = 0
Vì ( 1/1970 + 1/1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 -1/27 - 1/25 - 123 - 1/21 - 1/19 ) khác 0 nên để đẳng thức bằng 0 thì bắt buộc x - 1999 = 0
< = > x = 0 + 1999 = 1999
Vậy tập nghiệm của phương trình là S = { 1999 }
NQ
1
12 tháng 9 2019
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-163}{23}=10\)
\(\Leftrightarrow\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-166}{23}-4=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow x=258\)
Vậy \(x=258\)
Chúc bạn học tốt !!!
9 tháng 2 2022
\(\Leftrightarrow\dfrac{2032-x}{25}-1+\dfrac{2053-x}{23}-2+\dfrac{2070-x}{21}-3+\dfrac{2083-x}{19}-4=0\)
=>2007-x=0
hay x=2007
NA
9 tháng 2 2022
\(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}=10\)
\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)
\(\Leftrightarrow\dfrac{2032-x-25}{25}+\dfrac{2053-x-46}{23}+\dfrac{2070-x-63}{21}+\dfrac{2083-x-76}{19}=0\)
\(\Leftrightarrow\dfrac{2007-x}{25}+\dfrac{2007-x}{23}+\dfrac{2007-x}{21}+\dfrac{2007-x}{19}=0\)
\(\Leftrightarrow\left(2007-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)
\(\Leftrightarrow2007-x=0\left(vì.\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\ne0\right)\)
\(\Leftrightarrow x=2007\)
CH
Cô Hoàng Huyền
Admin
VIP
8 tháng 8 2016
\(pt\Leftrightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+...=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
Do \(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}>0\) nên 50 - x = 0 hay x = 50.
9 tháng 7 2017
pt<=>29-x/21+1+27-x/23+1+...=0
<=>50-x/21+50-x/23+50-x/25+50-x/27+50-x/29=0
<=>(50-x).(1/21+1/23+1/25+1/27+1/29)=0
Do 1/21+1/23+1/25+1/27+1/29>0 nên 50-x=0 hay x=50
17 tháng 1 2017
Ta có: \(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{23}=10\)
\(\Rightarrow\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{23}-10=0\)
\(\Leftrightarrow\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-170}{23}-4\right)=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow\left(x-258\right)=0\)
\(\Rightarrow x=258\)
Vậy x=258
\(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)
\(\Rightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)
\(\Rightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)
\(\Rightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Mà \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}>0\Rightarrow x-124=0\Rightarrow x=124\)